Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-6\right)^{2}$.
$$x^{2}-12x+36=-24x$$
Add $24x$ to both sides.
$$x^{2}-12x+36+24x=0$$
Combine $-12x$ and $24x$ to get $12x$.
$$x^{2}+12x+36=0$$
To solve the equation, factor $x^{2}+12x+36$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.
$$a+b=12$$ $$ab=36$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $36$.
Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x+6\right)\left(x+6\right)$$
Rewrite as a binomial square.
$$\left(x+6\right)^{2}$$
To find equation solution, solve $x+6=0$.
$$x=-6$$
Steps Using Factoring By Grouping
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-6\right)^{2}$.
$$x^{2}-12x+36=-24x$$
Add $24x$ to both sides.
$$x^{2}-12x+36+24x=0$$
Combine $-12x$ and $24x$ to get $12x$.
$$x^{2}+12x+36=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+36$. To find $a$ and $b$, set up a system to be solved.
$$a+b=12$$ $$ab=1\times 36=36$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $36$.
Rewrite $x^{2}+12x+36$ as $\left(x^{2}+6x\right)+\left(6x+36\right)$.
$$\left(x^{2}+6x\right)+\left(6x+36\right)$$
Factor out $x$ in the first and $6$ in the second group.
$$x\left(x+6\right)+6\left(x+6\right)$$
Factor out common term $x+6$ by using distributive property.
$$\left(x+6\right)\left(x+6\right)$$
Rewrite as a binomial square.
$$\left(x+6\right)^{2}$$
To find equation solution, solve $x+6=0$.
$$x=-6$$
Steps Using the Quadratic Formula
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-6\right)^{2}$.
$$x^{2}-12x+36=-24x$$
Add $24x$ to both sides.
$$x^{2}-12x+36+24x=0$$
Combine $-12x$ and $24x$ to get $12x$.
$$x^{2}+12x+36=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $12$ for $b$, and $36$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$x=\frac{-12±\sqrt{12^{2}-4\times 36}}{2}$$
Square $12$.
$$x=\frac{-12±\sqrt{144-4\times 36}}{2}$$
Multiply $-4$ times $36$.
$$x=\frac{-12±\sqrt{144-144}}{2}$$
Add $144$ to $-144$.
$$x=\frac{-12±\sqrt{0}}{2}$$
Take the square root of $0$.
$$x=-\frac{12}{2}$$
Divide $-12$ by $2$.
$$x=-6$$
Steps for Completing the Square
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-6\right)^{2}$.
$$x^{2}-12x+36=-24x$$
Add $24x$ to both sides.
$$x^{2}-12x+36+24x=0$$
Combine $-12x$ and $24x$ to get $12x$.
$$x^{2}+12x+36=0$$
Factor $x^{2}+12x+36$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x+6\right)^{2}=0$$
Take the square root of both sides of the equation.
$$\sqrt{\left(x+6\right)^{2}}=\sqrt{0}$$
Simplify.
$$x+6=0$$ $$x+6=0$$
Subtract $6$ from both sides of the equation.
$$x=-6$$ $$x=-6$$
The equation is now solved. Solutions are the same.
